Logically, I should have been able to do it this way, I thought, at least according to the docs. But instead it was giving me this error: TypeError: issubclass() arg 1 must be a class. (Uh, ok.) After a bit of searching, I eventually found this comment, and realized I must be trying to instantiate a form that the admin would instantiate automatically, had already been instantiated, or something like that pharmacieviagra.com.

After some more searching, I found the answer on the way to do what I needed here:

To add a link to an admin page on the user-side of your django site, you can use the named url patterns. (Seems like a simple and fairly straightforward thing to want to do, but I frequently find that things are not so simply found in most open source docs, nor are the examples always useful. But my usual open source docs rant is for another day. *g*)

It took a bit of digging, but I finally found that the admin url pattern options are listed here. From that, I was able to figure out that if I wanted to link to the edit (change action) page (parameter) for an article (model) in the app article, the pattern was:

{{ app_label }}_{{ model_name }}_action [parameters]

The namespace for the admin is, of course, admin, so putting that together using the {% url %} template tag, you get: